Mysql数据库中计算两GPS坐标的距离有两种方式：

1、直接使用SQL语句：#lat为纬度, lng为经度, 一定不要弄错

declare @lng1 float;

declare @lat1 float;

declare @lng2 float;

declare @lat2 float;

set @lng1=116.3899;

set @lat1=39.91578;

set @lng2=116.3904;

set @lat2=39.91576;

select (2*ATAN2(SQRT(SIN((@lat1-@lat2)*PI()/180/2)

*SIN((@lat1-@lat2)*PI()/180/2)+

COS(@lat2*PI()/180)*COS(@lat1*PI()/180)

*SIN((@lng1-@lng2)*PI()/180/2)

*SIN((@lng1-@lng2)*PI()/180/2)),

SQRT(1-SIN((@lat1-@lat2)*PI()/180/2)

*SIN((@lat1-@lat2)*PI()/180/2)

+COS(@lat2*PI()/180)*COS(@lat1*PI()/180)

*SIN((@lng1-@lng2)*PI()/180/2)

*SIN((@lng1-@lng2)*PI()/180/2))))*6378140;

#返回结果： 42.7484246368099

2、定义函数后再调用(适合批量计算)：# use aiezu;

drop function getDistance;

DELIMITER $$

CREATE DEFINER=`root`@`localhost` FUNCTION `getDistance`(

lng1 float(10,7)

,lat1 float(10,7)

,lng2 float(10,7)

,lat2 float(10,7)

) RETURNS double

begin

declare d double;

declare radius int;

set radius = 6378140; #假设地球为正球形，直径为6378140米

set d = (2*ATAN2(SQRT(SIN((lat1-lat2)*PI()/180/2)

*SIN((lat1-lat2)*PI()/180/2)+

COS(lat2*PI()/180)*COS(lat1*PI()/180)

*SIN((lng1-lng2)*PI()/180/2)

*SIN((lng1-lng2)*PI()/180/2)),

SQRT(1-SIN((lat1-lat2)*PI()/180/2)

*SIN((lat1-lat2)*PI()/180/2)

+COS(lat2*PI()/180)*COS(lat1*PI()/180)

*SIN((lng1-lng2)*PI()/180/2)

*SIN((lng1-lng2)*PI()/180/2))))*radius;

return d;

end

$$

DELIMITER ;

select getDistance(116.3899,39.91578,116.3904,39.91576); #调用函数

#返回结果：43.045058294389

常见问题及解决方法：

常见问题：

在创建函数前报如下错误：This function has none of DETERMINISTIC, NO SQL, or READS SQL DATA in its declaration and binary logging is enabled (you *might* want to use the less safe log_bin_trust_function_creators variable)解决方法：

在创建函数前执行下面SQL语句：set global log_bin_trust_function_creators=1;或者修改my.cnf文件并重启mysqld服务：log_bin_trust_function_creators=1

PHP计算两个GPS坐标点之间的距离: http://aiezu.com/article/18.html

Javascript计算两个GPS坐标的距离: http://aiezu.com/article/17.html

更多关于GPS计算请参考：http://www.movable-type.co.uk/scripts/latlong.html(被墙)